- #31
shawonna23
- 146
- 0
when using the sine law am I trying to find c?
Solving for Displacement in Circular Motion
- Thread starter shawonna23
- Start date
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- Displacement Physics
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Homework Help Overview
The problem involves calculating the displacement of a couple walking two-thirds of the way around a circular lake with a radius of 1.60 km. The couple starts at the west side of the lake and moves south, prompting questions about the magnitude and direction of their displacement relative to due east.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the method of determining displacement using trigonometric relationships and coordinate geometry. There are inquiries about the correctness of angles and the setup of the problem, including the use of diagrams and the sine law.
Discussion Status
Participants are actively engaged in exploring different methods to find the displacement. Some have provided guidance on drawing diagrams and using trigonometric functions, while others are questioning the assumptions made about angles and positions. There is a mix of confirmed and uncertain interpretations of the angles involved.
Contextual Notes
Some participants express confusion about the angles and the correct application of trigonometric principles. There is a noted urgency due to homework deadlines, and some constraints regarding the completeness of the information shared in the discussion.
- #32
shawonna23
- 146
- 0
Is it sin30/1.6=sin120/c
- #33
christinono
- 211
- 0
Here is the formula:
\frac{Sin A}{a} = \frac{Sin B}{b}
Now, place values in:
\frac{sin 30}{1.6km} = \frac{sin 120}{x}
isolate x:
x = \frac{sin 120}{sin 30} \times 1.6 km
Now, solve for x.
\frac{Sin A}{a} = \frac{Sin B}{b}
Now, place values in:
\frac{sin 30}{1.6km} = \frac{sin 120}{x}
isolate x:
x = \frac{sin 120}{sin 30} \times 1.6 km
Now, solve for x.
- #34
shawonna23
- 146
- 0
x=2.77 Now what do I do now?
- #35
christinono
- 211
- 0
What do you mean, that's the answer (I hope). Does you texbook or you worksheet give the answer?
- #36
MathStudent
- 281
- 1
thats the same answer i got 
...
...- #37
shawonna23
- 146
- 0
no it doesn't. So this is the answer for the magnitude. What is the direction in degrees?
- #38
christinono
- 211
- 0
Look back at the triangle.
- #39
shawonna23
- 146
- 0
Are you guys still there? What do I do to find the direction (relative to due east) in degrees?
- #40
shawonna23
- 146
- 0
Would I do this: theta= tan-1 (opp)/(adj)= 45 degrees?
- #41
christinono
- 211
- 0
Are U looking at you circle with the triangle in it? The angle U are looking for is the angle between the line that connects the 2 dots (initial and final positions of couple) and the line that connects the first dot to the origin.
- #42
christinono
- 211
- 0
Gotta run... :zzz: (actually, got to sleep)
BTW, the direction is 30 degrees North of East.
BTW, the direction is 30 degrees North of East.
- #43
shawonna23
- 146
- 0
thanks so much
- #44
christinono
- 211
- 0
Glad I could help. :!) 
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